JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 5)
Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :
$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$
$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$
$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$
$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$
Explanation
$$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$
$${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x $$
or $$f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$$
$$ = \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$$
$$f(x) = {{1 - x} \over {1 + x}}$$
Now, $$f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}$$
or $$f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}$$
or $${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$.
$${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x $$
or $$f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$$
$$ = \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$$
$$f(x) = {{1 - x} \over {1 + x}}$$
Now, $$f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}$$
or $$f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}$$
or $${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$.
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